Problem: Solve for $x$ and $y$ using elimination. ${6x+4y = 50}$ ${-3x+5y = 31}$
Answer: We can eliminate $x$ by adding the equations together when the $x$ coefficients have opposite signs. Multiply the bottom equation by $2$ ${6x+4y = 50}$ $-6x+10y = 62$ Add the top and bottom equations together. $14y = 112$ $\dfrac{14y}{{14}} = \dfrac{112}{{14}}$ ${y = 8}$ Now that you know ${y = 8}$ , plug it back into $\thinspace {6x+4y = 50}\thinspace$ to find $x$ ${6x + 4}{(8)}{= 50}$ $6x+32 = 50$ $6x+32{-32} = 50{-32}$ $6x = 18$ $\dfrac{6x}{{6}} = \dfrac{18}{{6}}$ ${x = 3}$ You can also plug ${y = 8}$ into $\thinspace {-3x+5y = 31}\thinspace$ and get the same answer for $x$ : ${-3x + 5}{(8)}{= 31}$ ${x = 3}$